A 68-kg water skier is being accelerated by a ski boat on a fl...

Question

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59).

(c) Explain why the skier's acceleration in part (b) is greater than that in part (a).

Accepted Answer

Hello, let's go through this practice problem. An industrial worker needs to move a 75 kg crate across a concrete floor. The coefficient of kinetic friction between the crate and the concrete floor is new. Sub K equals 0.31. The worker applies a pulling force of 255 new ones at an angle of six degrees from the horizontal to slide the crate. Considering the setup explain why the crate experiences a different acceleration when the force is applied at an angle compared to when the force is applied horizontally. A applying the force at an angle effectively reduces the crate's weight leading to less resistance and higher acceleration B applying the force at an angle increases the friction by pressing the crate more firmly against the surface purportedly resulting in reduced acceleration. C applying the force at an angle increases the horizontal component of the force leading to greater acceleration, irrespective of frictional force. And D applying the force at an angle slightly reduces the normal force which decreases the frictional force, opposing motion. This problem is essentially a conceptual question where if we think about two alternate scenarios for how this crate can be moved. So in one scenario, the rope is being pulled directly horizontally. And in another scenario, the force is being pulled at more of an angle with respect to the horizontal and rest to consider how and why these two different scenarios will result in different accelerations of the crate. And the answer to this becomes clear when we consider all of the other forces that are also acting on the crate aside from the force that the worker is applying. So on both crates, we're going to have a downward force from the weight of the crate itself. And as a result of the crate pressing down on the ground, there is also an upwards normal force on each of the crates and on both crates, there's also a frictional force opposing the direction of the applied force. So if you drew your applied force pointing to the right and the frictional force must be pointed to the left and vice versa. Now looking at our diagram, we can realize that if our goal is to move the crate horizontally, then the only force that is going to be impacting our ability to move the crate is the frictional force, which is the only other force shown in the diagram that is directed horizontally. If there's more friction, we can't accelerate the crate as much. If there's less friction, we can accelerate the crate more. Now recall that the frictional force is based on the magnitude of the normal force, whatever the frictional force is, that's acting at an object, it's equal to the coefficient of friction multiplied by the normal force that is being acted on by the object. And the normal force is something that has the ability to change. Recall that the normal force is reflective of how much the object is essentially being pushed up against the surface. Because as per Newton's third law, if, if one object is applying a force to another object, then that second object is applying a force of the same magnitude on the first object. So if we consider what's going on in the diagrams that we just drew notice that if we pull the box at an upwards angle, then even if the crate is still in contact with the ground, we're still going to be reducing some of the normal force that the ground itself is exerting on the box to think of it in more layman's terms. If it helps to think of it more intuitively, we're basically helping the box to not push so hard against the ground, which reduces the normal force. And if the normal force is reduced and the frictional force is also reduced as per the formula. So if we pull the box at an angle, then we're reducing the normal force, which is decreasing the frictional force that opposes the motion. And so what we just described is exactly what option D describes the other multiple choice options don't make sense. Option B kind of uses a similar line of logic but option B only makes sense if the angle is downwards who are pulling the box at an angle towards the floor. But the problem specifically states that this is an angle positive upwards from the horizontal. So it can't be B option A also doesn't make sense because unless we're fundamentally changing something about the crate itself, nothing we do is going to reduce its weight. And option C just doesn't make sense at all because if we apply the force at an angle that is going to decrease the horizontal component of the force, horizontal component is maximized if all of the force is in the horizontal direction. So options A through C don't make any sense and option C is the winner and that is it for this problem. I hope this video helped you out and please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now and I hope you have a lovely day. Bye bye.

Struggling with Physics?

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59).
<IMAGE>
(c) Explain why the skier's acceleration in part (b) is greater than that in part (a).

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Kinetic Friction practice set

Struggling with Physics?

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59).<IMAGE>(c) Explain why the skier's acceleration in part (b) is greater than that in part (a).

Watch next

Kinetic Friction practice set

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59).
<IMAGE>
(c) Explain why the skier's acceleration in part (b) is greater than that in part (a).